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1. Find the equation of the axis of symmetry of the graph of $y = (x + 3)^2 - 5$.

 

2. The x intercepts of the parabola y=2x^2+13x-7
are at (p,0)  and (q,0)  Find p and q

Enter your answer in the form "p, q ".

 

3. The x-intercepts of the parabola y=x^2+bx are  and  Find the equation of the parabola, and submit your answer in y=ax^2+bx+c form.

 Jan 16, 2021
 #1
avatar+116126 
+1

1.   The  vertex  is    (-3, -5)

The axis of symmetry   is   x  =  -3

 

2.   To find the x intercepts

 

2x^2 + 13x  -  7   =    0 

 

(2x -1)  ( x + 7)   =   0

 

Set  both factors to 0   and solve  for  x  and we  get that    x =   1/2 =  q     and  x  =   -7   =   p

 

3.   Something left out  here

 

 The x-intercepts of the parabola y=x^2+bx are  ??? and  ???

 

 

 

cool cool cool

 Jan 16, 2021
 #2
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im sorry for number 3 its The x-intercepts of the parabola y=x^2+bx are (-3,0)  and  (5,0) Find the equation of the parabola, and submit your answer in y=ax^2+bx+c form.

Guest Jan 16, 2021
 #3
avatar+116126 
+1

im sorry for number 3 its The x-intercepts of the parabola y=x^2+bx are (-3,0)  and  (5,0) Find the equation of the parabola, and submit your answer in y=ax^2+bx+c form.

 

 

The   x coordinate  of vertex  will   occur   halfway  between the x intercepts  =  (5 - 3) /2  =  1

 

And the x coordinate of the  vertex =   -b / (2a)       where a   = 1   

 

So

 

-b / 2  =  1

 

b  =  -2

 

And   since (5,0)  is on the  graphm we can solve for  c  thusly

 

0 =  5^2  - 2(5)  + c

0 = 15 + c

c =  -15

 

So  the equation  is

 

y = x^2  -2x  - 15

 

 

cool cool cool

 Jan 16, 2021

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